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I've been working on proving a conjecture I came up with a day or so ago but am having trouble getting off the ground. Here's the idea:

Define the trailing-ones number of a number $n\in \mathbb{N}$ to be the fewest number of 1's that must be added to the end of the number to make it prime.

For instance, the number $23$ has a trailing-ones number of $0$ because it is already prime without any ones being added to the end of it. On the other hand, the number $45$ has a trailing-ones number of $772$, because while $45$ is not prime, $45$ followed by 772 ones is a prime number. Moreover, any number of trailing ones $< 772$ after $45$ will not produce a prime number.

My conjecture is as follows:

Every natural number has a finite trailing-ones number.

That is, there exists no natural number that cannot be made prime by adding some finite number of 1's to the end of it.

What tools come to mind that I could use to prove this conjecture? Do you think this conjecture is false? If so, can you provide a counterexample? One particular example of note is that of $n=38$, which I have yet to compute the trailing-ones number of because it seems to be significantly greater than a thousand.

Bill Dubuque
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dsillman2000
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    An integer $n$ with $k$ digits, last of which is $1$, is a prime with probability inversely proportional to $\ln n$ which is more or less a constant times $k$. Because the harmonic series diverges, we would expect there to be infinitely primes in such a sequence. But, I'm not a number theorist. This was just a kind a heuristic. I've been told that surprisingly often primes do follow such heuristics, but proving it in this case is a much taller order :-) – Jyrki Lahtonen Mar 29 '22 at 21:29
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    This seems like an exceedingly difficult conjecture to prove. Very intriguing question. I do hope this gets upvotes so that more people see it. – Clyde Kertzer Mar 29 '22 at 21:41
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    In fact, $38$ is the smallest hard case. – Peter Mar 29 '22 at 22:54
  • @Peter Definitely is. 38 is very hard, but I've noticed something. Adding $3k+1$ ones always results in a multiple of 3, adding $3k+2$ ones always results in a multiple of 37, so we only need to think about adding $3k$ ones, no? EDIT: I believe 38 is actually a valid counterexample! Adding $3k$ ones is always a multiple of $23\cdots3$ (2 followed by $k$ 3's). – dsillman2000 Mar 29 '22 at 22:58
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    Etiquette on MSE would ask you to post your edit as an answer to your own question and then accept it. – Ian Mar 29 '22 at 23:14
  • @Peter in my edit I propose a covering system for $38$. Only missing rigor for the final case of $3k$ ones. – dsillman2000 Mar 29 '22 at 23:14
  • @Ian my counterexample is not yet totally proven for the case of $3k$ ones. But when it is, I'll follow your suggestion. – dsillman2000 Mar 29 '22 at 23:15
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    $$38111111111111111111111111111111111111111111111111111111111$$ has smallest prime factor $444007$ , but beginning with $371$ we seem to have always a very small prime factor. – Peter Mar 29 '22 at 23:16
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    Nice question & solution! – Jair Taylor Mar 29 '22 at 23:18
  • @Peter Two of the prime factors, 11661407 × 2000902063819 = 23333333333333333333 – dsillman2000 Mar 29 '22 at 23:18
  • @Ian No, updates should be made in the body of the question. Unless there is a complete solution for the problem , found by the author. – Peter Mar 29 '22 at 23:30
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    this example seems to exhibit algebraic factors – Peter Mar 30 '22 at 00:04
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    So, maybe $38$ is a "nontrivial" counterexample. – Peter Mar 30 '22 at 00:05
  • Counterexcamples follow easily by standard applications of covering arguments - see the linked dupe (and its links). Please search for answers before posing questions to help avoid duplication. – Bill Dubuque Mar 30 '22 at 01:16
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    Please don't put answers into the question - this is against SE policy. – Bill Dubuque Apr 07 '22 at 16:36

1 Answers1

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If we begin with $371$ , we never arrive at a prime :

  • If we add $6k$ ones ($k$ non-negative integer) , we have prime factor $7$
  • If we add $6k+3$ ones , we have prime factor $13$
  • If we add $3k+1$ ones , we have prime factor $3$
  • If we add $3k+2$ ones , we have prime factor $37$

Hence $[3,7,13,37]$ is a complete covering system.

Peter
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